For the previous example of the out of control bus which is forced to jump from a location $\langle 0,40,-5 \rangle$ with an initial velocity of $\langle 80,7,-5 \rangle$. We have now found the time of flight to be 9.59s and now want to find where the bus returns to the ground?

First find $v_{fy}$ for when it hits the ground. We need this in order to find $\vec{v_{avg}}$

$$ V_{fv} = V_{iy} + (\dfrac{F_{net,y}}{m}) \Delta{t}$$

$$ = V_{iy} + (\dfrac{-mg}{m}) \Delta{t}$$

$$ = V_{iy} - g\Delta{t}$$

$$ = 7m/s - (9.8 \dfrac{N}{kg})(9.59s)$$

$$ = -87m/s$$

Now to find the range:

$$ \vec{r_f} = \vec{r_i} + \vec{v_{avg}}\Delta{t}$$

$$ = \vec{r_i} + \dfrac{\vec{v_i} + \vec{v_f}}{2} \Delta{t}$$

$$ = \langle 0,40,-5 \rangle + (\dfrac{\langle 80,7,-5 \rangle m/s + \langle 80, -87, -5 \rangle m/s}{2})(9.59s)$$